Here's a simple answer (not necessarily the fastest):
overlapping[{a_, b_}, {c_, d_}] := Or[a == c, a < c < b, c < a < d]
Count[Flatten@Outer[overlapping, v1, v2, 1], True]
More efficient algorithms are possible if you can guarantee the bricks in
each list are non-overlapping and exhaustive.
Bobby
recode procedural algorithm to faster functional module
Subject: [mg44158] [mg44097] [mg44079] [mg44034] recode procedural algorithm to faster functional module
From: "Prince-Wright, Robert G SEPCO" <robert.prince-wright at shell.com>
To: mathgroup at smc.vnet.net
Can someone please help me recode this Module so it is less procedural
and hopefully runs a lot faster. The Lists V1 and V2 represent two time
series with 'bricks' laid out along them. The Bricks have varying length
set by v[[i]][[1]] and v[[i]][[2]] and the idea is to count the number
of times there is an overlap. I can only see the dumb procedural way of
taking each brick and checking if it overlaps in time with another!
Concurrence[v1_List, v2_List,nsim_]:=Module [
{k=1,c=0},
Do[
Do[
Which[
v1[[i]][[1]] == v2[[k]][[1]], c=c+1,
v1[[i]][[1]] < v2[[k]][[1]] && v1[[i]][[2]] > v2[[k]][[1]], c=c+1,
v1[[i]][[1]] > v2[[k]][[1]] && v1[[i]][[1]] < v2[[k]][[2]], c=c+1
];
(*Write[ strm1, {v1[[i]][[1]],v2[[i]][[1]],c}];*)
, {i,1,nsim}
], {k,1,nsim}
];
c
]
I've have a PowerPoint illustration if this is unclear and can e-mail it.
thanks
Robert Prince-Wright